# Example of product space isomorphic to sum of subspaces

Here is the problem statement (from chapter $$3$$ of Axler's Linear Algebra Done Right).

Give an example of a vector space $$V$$ and subspaces $$U_1,U_2$$ of $$V$$ such that $$U_1 \times U_2$$ is isomorphic to $$U_1 + U_2$$, but $$U_1 + U_2$$ is not a direct sum.

Hint: the vector space $$V$$ must be infinite-dimensional.

The only infinite-dimensional vector spaces mentioned in the book up to this point are $$\mathbb{F}^{\infty}$$ and $$P(\mathbb{R})$$.

I tried to construct an example using $$\mathbb{F}^{\infty}$$ by letting $$U_1$$ be the span of the standard bases $$\{e_{2n}\}$$ and $$U_2$$ the span of $$\{e_{2n-1}\}$$. It seems that at least one of the $$U_i$$ must be infinite dimensional, or else the example could be constructed with $$V$$ finite dimensional.

There is an isomorphism between $$U_1 \times U_2$$ and $$U_1 + U_2$$, but $$U_1 \cap U_2 = \{0\}$$ so we have a direct sum, which is what we're trying to avoid. I'm not sure how to proceed from here.

Modify your example to make the subspaces intersect. For example, we can take $U_1 = \operatorname{span} \{ e_1 \}$ and $U_2 = \operatorname{span} \{ e_i \}_{i \in \mathbb{N}}$. Then $U_1 \cap U_2 = U_1$ so $U_1 + U_2$ is not a direct sum. In addition, $U_1 \times U_2$ is isomorphic to $U_1 + U_2 = U_2$ via the linear map sending
$$(e_1,0) \mapsto e_1, (0,e_i) \mapsto e_{i + 1}.$$
• I didn't understand your linear map. However, I believe that it doesn't exist such vector space $V$. Because, if there's it, then the linear map $T$ is injective. But if $U_1 + U_2$ isn't a direct sum, then there's a non-zero $u \in U_1 \cap U_2$. So we have $T(u, -u) = 0$, which implies that T isn't injective. Contradiction. So there isn't such $V$. Apr 2, 2019 at 11:07
• @RafaelDeiga: It seems that you are assuming that the linear map $T \colon U_1 \times U_2 \rightarrow U_1 + U_2$ must be given by $T(u,v) = u + v$ so that $T(u,-u) = u - u = 0$. However, there's no reason $T$ must have this form. In my example, $e_1 \in U_1 \cap U_2$ but $T(e_1,0) = e_1, T(0,e_1) = e_2$ so $T(e_1,e_1) = T((e_1,0) + (0,e_1)) = T(e_1,0) + T(0,e_1) = e_1 + e_2 \neq 0$. Apr 2, 2019 at 11:16
• True. Indeed, I have assumed that $T$ is of the form $T(u,v) = u + v$. Now I understood your linear map. Apr 2, 2019 at 11:21